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Let $n\in\mathbb{N}$ fixed. How do you compute $$\sum_{n^2<m<(n+1)^2}\left\lfloor\frac{1}{\sqrt{m}-\lfloor\sqrt{m}\rfloor}\right\rfloor?$$

I did some examples for values of $n$, but didn't find a pattern.

Thanks in advanced

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    $\begingroup$ Even terms of oeis.org/A006218 . $\endgroup$ – Hw Chu May 25 '18 at 14:22
  • $\begingroup$ Thank you. Do yo have any idea to prove it? $\endgroup$ – sinbadh May 25 '18 at 15:29
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Write $m=n+k$ and rationalize the denominator to give $$\begin {align}\sum_{n^2<m<(n+1)^2}\left\lfloor\frac{1}{\sqrt{m}-\lfloor\sqrt{m}\rfloor}\right\rfloor&=\sum_{k=1}^{2n}\left\lfloor\frac{1}{\sqrt{n+k}-n}\right\rfloor\\&=\sum_{k=1}^{2n}\left\lfloor\frac{\sqrt{n+k}+n}{k}\right\rfloor\\&=\sum_{k=1}^{2n}\left\lfloor\frac{2n}{k}\right\rfloor \end {align}$$ showing that as Hw Chu comments this is the even numbered terms of OEIS A006218.

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